What Is the Relationship Between Voltage and Length in an Electrical Circuit?

[GreenPrint]

Homework Statement

So I was wondering if someone could help with part C. I have gotten the answers for A and B, at least I hope they are correct.

Homework Equations

The Attempt at a Solution

So I'll just do the rectangle first and then try and solve the trapezoid. By the way do you think it's a triangle or a trapezoid? To me it loos like a trapazoid with a value at point l=0 of (0,.1) and not (0,0)?

Anyways for the rectangle...

So I'm given

rho, electron concentration - n, the thickness of the graphite layer - h (found in part a), and the velocity - v (I found it), and the voltage across the rectangle is 2 volts.

I'm confused as to how to find the voltage across the length. I thought that the voltage across the length would be a constant 2 volts but I guess I'm wrong.

Thanks for any help anyone can provide.

 

[mfb]

I'm confused as to how to find the voltage across the length. I thought that the voltage across the length would be a constant 2 volts but I guess I'm wrong.

Voltage is 0 at the left side, and does not make jumps.
Current is constant, and resistance per length is constant. What can you tell about the voltage drop per length based on that?

 

[GreenPrint]

I'm not exactly sure. If you rearrange the formula

R = rho(l/A)

And in this case indeed

R/l = rho/A

I can tell that from resistance per length is constant

from ohms law

V = I R

divide both sides by L

V/L = (I R)/L

V/L = I (rho/A)

so then Voltage as a function of length...

V(L) = (L*I*rho)/A

Now for I... I just use the fact that the resistance is given, 100k ohms and use 2 volts for the voltage

V = I*R

I = V/R

in the equation to find voltage as a function of length?

 

[mfb]

Well, you can use a shortcut: Once you know that voltage is a linear function, you can directly get the result from the boundary conditions: 0V at the left side and 2V at the right side.

 

[GreenPrint]

http://img6.imageshack.us/img6/3449/capture6k.png

Ok I guess I'm doing something wrong as my answer to C for the voltage makes no sense. Thanks for any help. I have shown all steps.

For part (a)
R = 100 kΩ
l = 10 cm
ρ = $10^{-5}$ Ωm

R = $\frac{ρl}{A}$
$\frac{1}{R} = \frac{A}{ρl}$
A = $\frac{ρl}{R} = \frac{10^{-5} Ωm(10 cm)kΩ(100 cm)}{(100 kΩ)(1000 Ω) m} = 10^{-7} cm^{2}$

A = wh, $h = \frac{A}{w} = \frac{10^{-7}cm^{2}}{cm} = 10^{-7} cm$

For part (b) trapezoid only

R = $\frac{ρl}{A}$, $\frac{1-.1}{10}l + .1 = .09 l + .1 = w$, A = wh, A(l) = (.09l + .1)$10^{-7} cm^{2}$
R(l) = $\frac{ρl}{A(l)}, R = \frac{10^{-5} Ωm (100 cm)}{10^{-7}cm^{2}m} ∫^{10}_{0}\frac{l dl}{.09l + .1}dl ≈ 471.889 Ω$

I = nqvA, I = $\frac{V}{R}, \frac{V}{R} = nqvA, v = \frac{V}{RnqA} = \frac{2 V}{471.889 Ω\frac{10^{19}}{cm^{3}}(1.6*10^{-19})(.09l+.1)(10^{-7}cm^{2})} = \frac{2648866.285}{.09l+.1} \frac{cm}{s}$

V = IR, I = $\frac{V}{R} = \frac{2V}{471.889 Ω} ≈ 4.238*10^{-3} A$

R = $\frac{lρ}{A}, \frac{R}{l} = \frac{ρ}{A}, V = IR, \frac{V}{l} = \frac{IR}{l} = \frac{Iρ}{A}$
$V(l) = \frac{lIρ}{A} = \frac{l4.238*10^{-3} A(10^{-5} Ωm)(100 cm)}{(.09l+.1)10^{-7}cm^{2}m} = \frac{l42.38}{.09l + .1}$

 

[mfb]

The last two lines are problematic:
$\frac{dV}{dl}=\frac{I\rho}{A}$
It is the derivative here.
$V(l)=\int_0^l \frac{dV}{dl} dl' = \int_0^l \frac{I\rho}{A(l')} dl'$
Where A depends on l at the trapezoid.

 

[GreenPrint]

Wait so is the information you provide in your post correct?

V(l) = integral[0,l] [(I rho)/(A(l'))]dl'

 

[GreenPrint]

Ok so for the cross sectional area of the trapezoid...
dA = h * w(l) * dl

So

A(l) = 10^(-7) integral[0,l] [.09*l+.1]dl

 

[mfb]

Wait so is the information you provide in your post correct?

Do you have a reason to expect wrong formulas?

That (A in post 8) is not the same A you used above.
You don't need any integral to express A.

 

[GreenPrint]

Oh thanks for the help by the way. I really appreciate it =)

Oh I guess I thought you were asking if it was right or something.

So the formula I had for A originally was correct?

 

[GreenPrint]

So

$\frac{V}{l} = \frac{Iρ}{A(l)}$

$\frac{dV}{dl} = \frac{Iρ}{A}$

V(l) = $∫^{l}_{0} \frac{dV}{dl} dl = ∫^{l}_{0} dV = V$

I calculated earlier that

I = 4.238*$10^{-3} A$

and

A(l) = (.09*l + .1)$10^{-7}cm^{2}$

So

V(l) = $∫^{l}_{0} = \frac{4.238*10^{-3} A (10^{-5} Ωm (100 cm)}{(.09l + .1)10^{-7}cm^{2}m} dl$

I don't is right because when I plug in 10 I get about 423.8

 

[mfb]

Probably a numerical issue somewhere. Your resistance for the trapezoid is wrong - it has to be more than 100 kΩ (if you remove conductible material, the resistance increases), probably ~300-600 kΩ (471kΩ?)

 

[GreenPrint]

Ok thanks, I think I found my mistake and got the correct answer. I actually didn't get 471k for the resistance of the trapezoid? Is that what I'm supposed to get? I did find that my answer for the resistance for the trapezoid was indeed wrong but that's not what I got when I found the correct answer to that. Well I don't know if it's correct but when I plugged it into find the current and used that in my voltage equation and then plugged in 10 cm I got 2!

But my concern is that for the velocity of the square

I got something like

K/l

Where K is the constant that I found. If you plug in zero than the velocity is undefined. Is this a problem?

For the trapazoid I got something with a very well defined velocity at l = 0

I got something of the form

K/A(l)

When you plug in zero for L you don't get an undefined answer but a very well defined one. Is this an issue?

 

[mfb]

471kΩ was just a guess, based on your previous value of 471Ω. I did not calculate it.

The velocity in the square should be the same everywhere, you can calculate it with the known cross-section area, current and material constants, and they are all independent of l.

 

[GreenPrint]

Thanks for all the help. I think I got the right answer (because of your help) =)